The top of a waterfall is 948 m high - NSC Technical Sciences - Question 4 - 2022 - Paper 1

To calculate the gravitational potential energy, we use the formula:

$E_p = mgh$

Where:

• $m = 6 imes 10^4 \, kg$ (mass of the water)
• $g = 9.81 \, m/s^2$ (acceleration due to gravity)
• $h = 948 \, m$ (height of the waterfall)

Substituting these values, we have:

E_p \approx 5.57 \times 10^6 \; J$$

At the highest point, the mechanical energy (M_e) consists only of gravitational potential energy:

$M_e = mgh$

Where:

• $m = 0.22 \, kg$ (mass of the pendulum bob)
• $h = 0.25 \, m$ (height above the floor)

Substituting these values, we calculate:

$M_e = 0.22 \, kg \times 9.81 \, m/s^2 \times 0.25 \, m \approx 0.539 \; J$

At the lowest point, the mechanical energy (M_e) consists of both kinetic energy (K) and potential energy (E_p), but since the height is zero, the potential energy is:

$M_e = K + E_p$

This results to $E_p = 0$ at the lowest point, thus:

$M_e = K$

At the lowest point, we can use the conservation of energy principle:

$E_{p_{ ext{top}}} = E_{k_{ ext{bottom}}}$

Setting this up:

$M_e = \frac{1}{2} mv^2$

From 4.3.1, we know:

$M_e = 0.539 \, J$

Now we can solve for $v$:

$0.539 \, J = \frac{1}{2} \times 0.22 \, kg \times v^2$

Solving for $v$ gives:

$v^2 = \frac{2 \times 0.539}{0.22} \approx 4.9$

Thus,

$v \approx \sqrt{4.9 \; m^2/s^2} \approx 2.21 \, m/s$